A082369 Numbers congruent to 13 mod 30.

13, 43, 73, 103, 133, 163, 193, 223, 253, 283, 313, 343, 373, 403, 433, 463, 493, 523, 553, 583, 613, 643, 673, 703, 733, 763, 793, 823, 853, 883, 913, 943, 973, 1003, 1033, 1063, 1093, 1123, 1153, 1183, 1213, 1243, 1273, 1303, 1333, 1363, 1393, 1423, 1453

Offset: 1

Author

Cino Hilliard, May 11 2003

Keywords

• nonn
• easy

Comments

Solutions to 19^x + 23^x == 29 mod 31.

The form of these numbers is obviously 30X + 13. 3^x + 5^x == 7 mod 11 and 17^x + 19^x == 23 mod 29 have no solutions. In fact, 3^x + 5^x == m mod 11 is only solvable for m < 11 = 1, 2, 8, 9. Similarly, 17^x + 19^x == m mod 29 is not solvable for m < 29 = 6, 11, 13, 14, 15, 16, 18, 23. I can't even prove 3^x + 5^x-7 <> 11k for all integers x, k. Anyone have a general proof of these statements say, a^x + -b^x == m mod k true or false for certain a, b, m, k, x combinations?

a^x + b^x == m (mod k) is periodic mod phi(k), so it suffices to check x = 1, 2, ..., phi(k). - Charles R Greathouse IV, Nov 19 2013

Links

• Michael G. Kaarhus, Table of n, a(n) for n = 1..10000
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Programs

Mathematica

Range[13, 7000, 30] (* Vladimir Joseph Stephan Orlovsky, Jul 13 2011 *)
LinearRecurrence[{2,-1},{13,43},50] (* Harvey P. Dale, Mar 02 2023 *)

Maxima

f(a):= mod((19^a + 23^a),31)$ a:-1$ for n:1 thru 3000 step 0 do(a:a+1, if f(a)=29 then (if mod(a,30)=13 then (print(n," ",a), n:n+1) else (print("Exception at ",a,", ",f(a)), n:3001))); /* f(a)==29 only when a is cong. to 13 (mod 30). No exceptions thru a=89983, n=3000. Michael G. Kaarhus, Nov 18 2013 */

PARI

anpbn(n)= for(x=1,n, if((19^x+23^x-29)%31==0,print1(x, ", "))) \\ solutions to 19^x+23^x == 29 mod 31

Formula

a(n) = 30n + 13.

G.f.: x*(13+17*x)/(1-x)^2. - Colin Barker, Jan 11 2012

Extensions

Simpler name from Charles R Greathouse IV, Nov 19 2013